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Week 1, cos 441, fall 2008
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-- Continue examples not covered from Lecture 1

-- Go over exercise 1


Rule Induction
--------------

-- the given inference rules are an *exhaustive* source of information
about the valid judgments.
-- no judgment is valid unless you can derive a proof for it using 
some finite number of the given inference rules

-- this leads to a proof principal (Rule Induction):

   * to show every derivable judgment has some property P, show for every
     rule in the deductive system:

      J1 ... Jn
      --------- name
          J

     if J1 ... Jn have property P then J has property P

-- examples:

1. Given a property P, we know that P is true for all natural numbers, 
if we can prove:

   * P holds unconditionally for Z.  Corresponds to rule:

                ----------- Z        
                 |- Z nat                

   * Assuming P holds for n then P holds for (S n).  Corresponds to rule:

                   |- n nat
                ------------- S
                  |- S n nat

-- also called "induction on the structure of natural numbers" or more 
generally "structural induction" when the objects are numbers, but other 
things like trees or programming language expressions.


Theorem:  If |- even2 n then |- even n.
Proof:
By induction on the derivation |- even2 n.

case:

---------- even2Z         
|- even2 Z           

|- even Z             (by evenZ)

case:

|- even2 n       
-----------------  even2S 
|- even2 (S (S n))


|- even n            (by IH)
|- odd (S n)         (by oddS)
|- even (S (S n))     (by evenS)

QED



Theorem:  If |- n nat then either |- even n  or |- odd n.

Proof:

By induction on the derivation of |- n nat.  
(Or "by induction on the structure of naturals n")

Case:  

--------
|- Z nat 


|- even Z     (by rule evenZ)


case:

 |- n nat
-------------
 |- S n nat


(1) |- even n    or   (2) |- odd n        (by IH)

Assuming (1):
|- odd S n                                (by (1) and oddS)

Assuming (2):
|- even S n                               (by (2) and evenS)

QED


Theorem:  For all n1, n2, there exists n3 such that |- add n1 n2 n3.

Proof:
By induction on the structure of n1.

case:

--------
|- Z nat 


|- add Z n2 n2     (by addZ)

case:

 |- n nat
------------- succ
 |- S n nat


|- add n n2 n3'        (by IH)
|- add (S n) n2 (S n3')  (by addS)

QED